# Question #10e47

##### 2 Answers
Jan 18, 2018

Toque applied = $6 N . m$ & power is about$120$ watt

#### Explanation:

Here,by applying the torque angular retardation was done to stop its motion(considering no skidding was there).

so, angular retardation= $\alpha$ = change in angular velocity i.e $\frac{\omega}{t}$ = $\frac{40}{10}$ i.e $4$ radian/${s}^{2}$

Now, torque =$I \alpha$ (where, I is the moment of inertia)

moment of inertia of a cylinder passing through its axis is $\frac{M {r}^{2}}{2}$ =$1.5 K g . {m}^{2}$

so. $\tau$ =$1.5 \cdot 4 N . m$ i,e $6 N . m$

so,angular power required = $\frac{\tau \cdot \theta}{t}$ i.e about $120$ watt($\tau \cdot \theta$= work done,dividing by time required for this angular displacement of $\theta$ by the time $t$ in which it was brought to rest due to application of this torque) ( $\theta$ is angular displacement)($\theta = {\left(\omega\right)}^{2} / \left(2 \alpha\right) = \frac{{40}^{2}}{2 \cdot 4} = 200 r a \mathrm{di} a n$)

Jan 18, 2018

The torque is $= 6 N m$. The power is $= 240 W$

#### Explanation:

The torque and the angular acceleration are related

$\text{Torque (Nm)"="Moment of inertia (kgm^2)"xx "angular acceleration (rads^-2)}$

$\tau = I \times \alpha$

The moment of inertia of the solid cylinder is

$I = \frac{m {r}^{2}}{2}$

The mass of the cylinder is $m = 3 k g$

The radius of the cylinder is $r = 1.0 m$

The moment of inertia is $I = \frac{3 \times {1}^{2}}{2} = 1.5 k g {m}^{2}$

The initial angular velocity is ${\omega}_{0} = 40 r a {\mathrm{ds}}^{-} 1$

The final angular velocity is ${\omega}_{1} = 0 r a {\mathrm{ds}}^{-} 1$

The time is $t = 10 s$

Apply the equation (to find the angular acceleration $\left(\alpha\right)$)

${\omega}_{1} = {\omega}_{0} + \alpha t$

$\alpha = \frac{{\omega}_{1} - {\omega}_{0}}{t} = \frac{0 - 40}{10} = - 4 r a {\mathrm{ds}}^{-} 2$

The torque to be applied is

$\tau = I \alpha = 1.5 \times 4 = 6 N m$

$\text{Power (W)"= "torque (Nm)"xx "angular velocity (rads^-1)}$

$P = \tau \times {\omega}_{0}$

The power is $P = 6 \times 40 = 240 W$